Let there are 3 bags A, B and C. Bag contain 5 black balls and 7 red balls, bag B contains 5 red and 7 black balls and bag C contains 7 red and 7 black balls. A balls is drawn and found to be black, then the probability that it is drawn from bag A, is

Options

A. `7/(18)`
B. `5/(42)`
C. `5/(18)`
D. `1/3`

Answer

To solve the problem, we need to find the probability that a black ball drawn is from bag A, given that a black ball has been drawn. We will use Bayes' theorem for this calculation.

### Step 1: Define the events
Let:
- \( A \): Event that the ball is drawn from bag A.
- \( B \): Event that the ball is drawn from bag B.
- \( C \): Event that the ball is drawn from bag C.
- \( E \): Event that a black ball is drawn.

### Step 2: Calculate the probabilities of drawing a black ball from each bag
1. **Bag A**: Contains 5 black balls and 7 red balls (total 12 balls).
- Probability of drawing a black ball from bag A:
\[ P(E|A) = \frac{5}{12} \]

2. **Bag B**: Contains 7 black balls and 5 red balls (total 12 balls).
- Probability of drawing a black ball from bag B:
\[ P(E|B) = \frac{7}{12} \]

3. **Bag C**: Contains 7 black balls and 7 red balls (total 14 balls).
- Probability of drawing a black ball from bag C:
\[ P(E|C) = \frac{7}{14} = \frac{1}{2} \]

### Step 3: Calculate the prior probabilities of choosing each bag
Since there are three bags and each bag is equally likely to be chosen:
\[ P(A) = P(B) = P(C) = \frac{1}{3} \]

### Step 4: Calculate the total probability of drawing a black ball
Using the law of total probability:
\[ P(E) = P(E|A)P(A) + P(E|B)P(B) + P(E|C)P(C) \]
Substituting the values:
\[ P(E) = \left(\frac{5}{12} \cdot \frac{1}{3}\right) + \left(\frac{7}{12} \cdot \frac{1}{3}\right) + \left(\frac{1}{2} \cdot \frac{1}{3}\right) \]
Calculating each term:
\[ P(E) = \frac{5}{36} + \frac{7}{36} + \frac{6}{36} = \frac{18}{36} = \frac{1}{2} \]

### Step 5: Apply Bayes' theorem to find the probability that the black ball is from bag A
Using Bayes' theorem:
\[ P(A|E) = \frac{P(E|A)P(A)}{P(E)} \]
Substituting the known values:
\[ P(A|E) = \frac{\left(\frac{5}{12}\right) \left(\frac{1}{3}\right)}{\frac{1}{2}} = \frac{\frac{5}{36}}{\frac{1}{2}} = \frac{5}{36} \cdot \frac{2}{1} = \frac{10}{36} = \frac{5}{18} \]

### Final Answer
The probability that the black ball drawn is from bag A is:
\[ \boxed{\frac{5}{18}} \]